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The equation of the normal to the circle $(x-1)^2+(y-2)^2=4$ which is at a maximum distance from the point $(-1,-1)$ is

(A) $x+2y=5$
(B) $2x+y=4$
(C) $3x+2y=7$
(D) $2x+3y=8$

Since its a normal to the circle at max distance from the point, I took any general point on the given circle and by using the distance formula I tried to use maxima minima concept....but I am not getting the answer...... Moreover every normal must pass through the centre of the circle also..... Please help me out........

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  • $\begingroup$ Hint: The normal (to the circle) whose distance to the point $(-1,-1)$ is a maximum must be perpendicular to the line connecting that point to the center of the circle. (Why?) $\endgroup$ – hardmath Feb 10 '16 at 16:18
  • $\begingroup$ Is this a homework question ? What have you tried so far ? $\endgroup$ – Shubham Marathia Feb 10 '16 at 16:20
  • $\begingroup$ Well one very easy way would be just to find out the distance of $(-1,-1)$ from the given lines in the options. If this was a question from a competitive exam that's what you should do. $\endgroup$ – GTX OC Feb 10 '16 at 16:34
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The centre of the circle is $(1,2)$. The normal by its nature must go through $(1,2)$.

Since it must be as far from $(-1,-1)$ as possible, the slope of the normal will be perpendicular to the line joining $(-1,-1)$ and $(1,2)$.

The slope of the line joining $(-1,-1)$ and $(1,2)$ is $3/2$, so the slope of the normal is $-2/3$. We now know the slope of the normal, and a point it goes through (i.e. $(1,2)$), so we can find the equation using $y - y_1 = m(x - x_1)$.

It turns out to be $2x + 3y = 8$, so (D)

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  • $\begingroup$ Bro even I did the same but the is not D $\endgroup$ – Aditya Feb 10 '16 at 16:20
  • $\begingroup$ "by definition" is wrong. The fact that the lines normal to the circle are precisely those that pass through the center is a demonstrable proposition, not a definition. $\qquad$ $\endgroup$ – Michael Hardy Feb 10 '16 at 16:21
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    $\begingroup$ I'd say "perpendicular to the line" rather than "perpendicular to the slope of the line". $\qquad$ $\endgroup$ – Michael Hardy Feb 10 '16 at 16:21
  • $\begingroup$ Answer given is option A $\endgroup$ – Aditya Feb 10 '16 at 16:23
  • $\begingroup$ Even I don't know why...... $\endgroup$ – Aditya Feb 10 '16 at 16:23
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The circle $(x- 1)^2+ (y- 2)^2= 4$ has center (1, 2). The points on that circle closest to and farthest from (-1, -1) lie on the line from (-1, -1) to (1, 2). That is the line $y= (-3/2)(x+ 1)- 1= (-3/2)x- 5/2$. The two points on the circle are the two (x, y) values you get by solving $(x- 1)^2+ (y- 2)^2= 4$ and $y= (-3/2)x- 5/2$ simultaneously. The simplest way to do that is to replace y in the quadratic equation by $(-3/2)x- 5/2$ to get a single quadratic equation in x.

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